GATE CSE 2024 | Set 1 | Question: 8

Question

Consider the following $\mathrm{C}$ program:

#include <stdio.h>

int main() {

int a=6;

int b = 0;

while (a<10)   {

a = a / 12+1 ;

a += b ;}

printf ("%d", a);

return 0 ; }

Which one of the following statements is CORRECT?

• A. The program prints $9$ as output
• B. The program prints $10$ as output
• C. The program gets stuck in an infinite loop
• D. The program prints $6$ as output

Answer 1

The initial value is $a=6,b=0$. When the first time while loop is runs the conditions become true as $(6<10)$. Now control enters inside the for loops and executes the following lines:

$a=a/12+1\implies (6/12)+1=0+1=1$: it returns $a=1$

$a=a+b \implies1+0=1$: it returns again $a=1$

Now in the second iteration again while conditions become true as $(1<10)$ and again control enters inside the loops and executes the inside statements:

$a=a/12+1$: it returns $a=1$

$a=a+b$: it returns again $a=1$

If we execute again and again every time conditions become true. So no output is printed and programs go into infinite loops.

Answer 2

In 1st iteration,

a = 6/10 +1
a += 0
Since a is an integer, a = 1 at the end of first iteration

In 2nd iteration,

a = 1/10 +1
a += 0
Since a is an integer, a = 1 at the end of second iteration

So, value of a will always be less than 10, the while loop will never end.

Answer: option C

Answer 3

Inside the while loop a is a dependent variable that while loop will use for it's recurrence.

and we have a as,

a = a/12 + 1;

a += b; 

with a = 6 and b = 0 as initial values.

so, a = 6/12 + 1 = 0.5 + 1 = 0 + 1 [because a is an integer variable where any values after decimal point are simply dropped.]

and then a = 1 + 0 = 1

similarly more runs of a will make the value of a as 1 only. Because of the integer division.

Hence the program will never end. Option C.